The lifespans of meerkats in a particular zoo are normally distributed. The average meerkat lives $15.3$ years; the standard deviation is $3.3$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a meerkat living between $5.4$ and $8.7$ years.
Answer: $15.3$ $12$ $18.6$ $8.7$ $21.9$ $5.4$ $25.2$ $99.7\%$ $95\%$ $2.35\%$ $2.35\%$ We know the lifespans are normally distributed with an average lifespan of $15.3$ years. We know the standard deviation is $3.3$ years, so one standard deviation below the mean is $12$ years and one standard deviation above the mean is $18.6$ years. Two standard deviations below the mean is $8.7$ years and two standard deviations above the mean is $21.9$ years. Three standard deviations below the mean is $5.4$ years and three standard deviations above the mean is $25.2$ years. We are interested in the probability of a meerkat living between $5.4$ and $8.7$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $99.7\%$ of the meerkats will have lifespans within 3 standard deviations of the average lifespan. It also tells us that $95\%$ of the meerkats will have lifespans within 2 standard deviations of the mean. That leaves $99.7\% - 95\% = 4.7\%$ of meerkats between 2 and 3 standard deviations of the mean, or $2.35\%$ on either side of the distribution. The probability of a particular meerkat living between $5.4$ and $8.7$ years is $\color{orange}{2.35\%}$.